Electric–magnetic duality from familiar gauge theory

In source-free Maxwell theory, the equations for electric and magnetic fields can be exchanged. A quantum theory asks a sharper question: does that exchange preserve the allowed fluxes and their energies, with the coupling transformed appropriately?

We will answer this question in a finite box before discussing interacting Yang–Mills theory. The preparation is the complex coupling and conformal family and compact connections, flux and holonomy. No Seiberg–Witten periods or superspace calculation is needed.

The eventual interacting theory is four-dimensional $\mathcal N=4$ Yang–Mills with gauge algebra $\mathfrak{su}(N)$, $N\geq2$, and sixteen real Poincaré supercharges. Its $\mathcal N=1$ presentation displays four. The initial calculation is instead an ordinary compact $U(1)$ Maxwell field. It uses no supersymmetry and makes no approximation in rank or coupling. Its flux eigenstates are not being asserted to form a BPS spectrum.

Fix the electric unit before exchanging fields

Let $a$ be a compact connection whose unit Wilson probe is $\exp(i\oint a)$. The probe need not be a dynamical particle. Its curvature $f=da$ has magnetic periods

$$\frac1{2\pi}\int_\Sigma f\in\mathbb Z. \tag{N17.1}$$

For signature $(+---)$ and $\epsilon_{0123}=+1$, define $E_i=f_{0i}$ and $B_i=\epsilon_{ijk}f_{jk}/2$, with $\epsilon_{123}=+1$. We fix both the kinetic coefficient and theta sign by

$$\mathcal L=\frac{\mathbf E^2-\mathbf B^2}{2e^2} -\frac{\theta}{4\pi^2}\mathbf E\cdot\mathbf B, \qquad \tau_A=s+it,\quad s=\frac{\theta}{2\pi},\quad t=\frac{2\pi}{e^2}>0. \tag{N17.2}$$

Here $[a]=1$, $[E]=[B]=2$ and $e,\theta,\tau_A$ are dimensionless. The subscript distinguishes this explicitly normalized Abelian example from a coupling imported without its charge convention.

Why is there $2\pi/e^2$, rather than $4\pi/g^2$? For a one-dimensional ordinary trace, the previous gauge-action normalization is reproduced by $e^2=g^2/2$. Then $\tau_A=\theta/(2\pi)+4\pi i/g^2$. A non-Abelian Cartan generator with trace norm $1/2$ has a different component coefficient; its allowed fluxes must be translated too. A coupling formula alone does not determine that translation.

Quantize the fluxes in a box

Take space to be a rectangular three-torus with side lengths $L_i$ and volume $\mathcal V=L_1L_2L_3$. In a sector with magnetic integers $m_i$, a uniform representative has

$$B_i=\frac{2\pi m_i L_i}{\mathcal V}. \tag{N17.3}$$

For nonzero flux the connection is patched, not a single global one-form. Nevertheless, the difference between it and a fixed reference connection on the same bundle is a global one-form. Add a spatially constant flat part and denote its three angular holonomies by $q_i$. Large gauge transformations identify $q_i\sim q_i+2\pi$. In temporal gauge, its electric field is $E_i=\dot q_i/L_i$.

Substituting these fields into (N17.2) and integrating over space gives the mechanical Lagrangian

$$L_0=\sum_i\left[ \frac{\mathcal V}{2e^2L_i^2}\dot q_i^2 -s m_i\dot q_i -\frac{2\pi^2L_i^2}{e^2\mathcal V}m_i^2 \right]. \tag{N17.4}$$

The theta term is a total time derivative in a fixed magnetic sector. It still changes canonical momentum:

$$p_i=\frac{\partial L_0}{\partial\dot q_i} =\frac{\mathcal V}{e^2L_i^2}\dot q_i-sm_i. \tag{N17.5}$$

A wavefunction on the holonomy circle is periodic. Thus $p_i=-i\partial_{q_i}$ has integer eigenvalues $n_i$, carried by $\exp(i n_iq_i)$. The electric field is proportional to $n_i+s m_i$, not just to $n_i$. This is the elementary Hamiltonian origin of the theta-dependent electric shift.

The Legendre transform cancels the explicit term linear in velocity, but the relation between velocity and momentum retains it:

$$\begin{aligned} H_0 &=\sum_i\frac{e^2L_i^2}{2\mathcal V} \left[(n_i+s m_i)^2+t^2m_i^2\right]\\ &=\sum_i\frac{\pi L_i^2}{\mathcal V} \frac{|n_i+\tau_A m_i|^2}{\operatorname{Im}\tau_A}. \end{aligned} \tag{N17.6}$$

This expression is positive and has dimension one. At $\theta=0$, a pure electric flux costs $e^2L_i^2n_i^2/(2\mathcal V)$, while a pure magnetic flux costs $2\pi^2L_i^2m_i^2/(e^2\mathcal V)$. Weak electric coupling makes the former cheap and the latter expensive.

The integers $(n_i,m_i)$ label flux sectors through a torus, not charges of particles that we have constructed. Nonzero-momentum photons form additional oscillator modes. Equation (N17.6) is their separate uniform-flux contribution, exact for the free Maxwell theory.

Exchange the coupling and the labels together

Consider one of the three charge pairs and suppress its spatial index. Define

$$S:\qquad \tau_A'=-\frac1{\tau_A},\qquad (n',m')=(m,-n). \tag{N17.7}$$

Direct substitution, including the imaginary part, gives

$$n'+\tau_A'm'=\frac{n+\tau_A m}{\tau_A}, \qquad \operatorname{Im}\tau_A' =\frac{\operatorname{Im}\tau_A}{|\tau_A|^2}. \tag{N17.8}$$

The numerator and denominator of (N17.6) acquire the same factor. Every flux energy is preserved, and the map is a bijection of $\mathbb Z^2$. Applying $S$ twice reverses both fluxes: $S^2(n,m)=(-n,-m)$. It is charge reversal, rather than the identity on oriented charge labels.

For a worked example, let the box be cubic with side $L$, and excite only the first pair:

$$\tau_A=\frac{1+3i}{2},\quad(n,m)=(2,-1), \qquad \tau_A'=\frac{-1+3i}{5},\quad(n',m')=(-1,-2). \tag{N17.9}$$

Before the exchange, $|2-\tau_A|^2/(3/2)=3$. After it, $|-1-2\tau_A'|^2/(3/5)=3$. Both descriptions give $H_0=3\pi/L$. Keeping the original integers while changing only $\tau_A$ would not perform this test.

At $\theta=0$, the same map gives $e_D=2\pi/e$, or $g_D=4\pi/g$ in the one-dimensional trace translation above. At nonzero theta, both the real and imaginary parts transform; replacing $e$ by $2\pi/e$ alone is then incorrect.

Separate a theta shift from a change of units

The other useful transformation is

$$T:\qquad \tau_A'=\tau_A+1,\qquad (n',m')=(n-m,m). \tag{N17.10}$$

This is a relabeling of the same flux energy: $n'+\tau_A'm'=n+\tau_A m$. If instead the integer $n$ is held fixed while theta is increased by $2\pi$, the electric field label $n+s m$ increases by $m$. These are different comparisons. The latter is the Witten-effect shift; the inverse relabeling in (N17.10) identifies spectra at the two endpoints. This distinction also explains the plus-sign electric-weight shift in Aharony–Seiberg–Tachikawa, (2.3). This calculation is on the spin spacetime $\mathbb R\times T^3$; it does not establish the same theta periodicity for arbitrary non-spin four-manifolds.

More generally, let $a_0,b_0,c_0,d_0$ be integers with $a_0d_0-b_0c_0=1$. The set of such matrices is called $SL(2,\mathbb Z)$. The corresponding flux identification is

$$\begin{gathered} \tau_A'=\frac{a_0\tau_A+b_0}{c_0\tau_A+d_0}, \qquad n'=a_0n-b_0m,\quad m'=-c_0n+d_0m,\\ n'+\tau_A'm'=\frac{n+\tau_A m}{c_0\tau_A+d_0}, \qquad t'=\frac{t}{|c_0\tau_A+d_0|^2}. \end{gathered} \tag{N17.11}$$

The subscripted matrix entries are numbers, unrelated to the connection $a$. The integer determinant-one condition makes the charge map invertible over the integers. It also preserves the antisymmetric pairing of two labels,

$$\langle(n_1,m_1),(n_2,m_2)\rangle =n_1m_2-m_1n_2. \tag{N17.12}$$

For electric and magnetic probes, this is the integral Dirac pairing in unit-charge conventions. Preserving a continuous quadratic expression without preserving the allowed integral labels would not suffice.

One further check sums rather than follows the flux states. For one spatial direction and $b=\beta\pi L_i^2/\mathcal V>0$, define

$$Z_{\rm flux}(b,\tau_A) =\sum_{n,m\in\mathbb Z} \exp\!\left[-\frac b t|n+\tau_A m|^2\right]. \tag{N17.13}$$

Here $\beta>0$ has dimension of inverse energy. This is just a positive flux trace, not the full supersymmetric partition function. Its invariance follows by the bijections above. An independent evaluation follows by Fourier transforming a Gaussian in $n$:

$$Z_{\rm flux} =\sqrt{\frac{\pi t}{b}}\sum_{w,m\in\mathbb Z} \exp\!\left[-btm^2-\frac{\pi^2t}{b}w^2\right] e^{2\pi i wsm}. \tag{N17.14}$$

The nonzero-theta phases make this a useful normalization check. The two sums agree with controlled Gaussian tails. No conclusion about curvature-dependent partition-function factors on other four-manifolds follows from this flat flux trace.

What the non-Abelian statement adds

The complex coupling of $\mathcal N=4$ Yang–Mills labels an exactly marginal family. A duality map relates descriptions at different values of that parameter; it is not evolution with the renormalization scale. The non-Abelian $SL(2,\mathbb Z)$ claim is an additional nonperturbative input, explicitly distinguished from conformality in Minahan’s introduction. The Maxwell calculation does not prove it.

For $\mathfrak{su}(N)$ the electric and magnetic Lie algebras agree, but their global gauge groups and permitted line probes need not. A line probe can specify electric holonomy, magnetic flux around the line, or both; it need not describe an elementary particle. The duality must map this full set as well as the coupling. These are the data emphasized in Aharony–Seiberg–Tachikawa, §1.1 and (2.10)–(2.12).

Thus the appropriate statement concerns a family of theories with specified global data. It does not make every fixed $SU(N)$ theory invariant under every matrix in (N17.11). Nor may $U(N)$ be silently replaced by $SU(N)$: the former has a locally free identity multiplet, and its global group is $(SU(N)\times U(1))/\mathbb Z_N$, rather than a direct product.

On vacua where commuting scalar expectation values break the non-Abelian group to Abelian factors, photon variables are a useful local description. That observation still does not identify their charge lattices with the unit $U(1)$ example by inspection. The prepotential dualization lesson develops a different, explicitly normalized N=2 application; it is optional here.

Next, derive the global gauge groups and line spectra that accompany a chosen gauge algebra and follow their duality maps. Carry the electric and magnetic units with you: the compact Maxwell labels above cannot be copied unchanged into a non-Abelian Cartan lattice.

Exercises

1. Follow a magnetic sector through a theta cycle

In a cubic box of side $L$, hold $e$ fixed and restrict to $m_1=1$, $m_2=m_3=0$. Find the lowest energy within this magnetic sector as $s=\theta/(2\pi)$ runs from zero to one. Identify the electric integers at the crossing, and compare tracking a fixed state with choosing the lowest state at each theta.

Hint 1

Minimize the square $(n_1+s)^2$ over integers. The zero-flux electric labels in the other directions minimize their own squares.

Hint 2

The nearest integer to $-s$ changes at $s=1/2$. Compare the endpoint labels using (N17.10).

Solution

Set $n_2=n_3=0$. For $0\leq s\leq1/2$, the minimizing label is $n_1=0$; for $1/2\leq s\leq1$, it is $n_1=-1$. At the midpoint both states are present and degenerate. Therefore

$$E_{\min}(s) =\frac{e^2}{2L} \left[t^2+\min\{s^2,(1-s)^2\}\right].$$

At the crossing the energy is $e^2(t^2+1/4)/(2L)$; at either endpoint it is $e^2t^2/(2L)$. A state kept at $n_1=0$ instead follows $e^2(t^2+s^2)/(2L)$ throughout and is not the minimum for $s>1/2$.

The $T$ relabeling sends the initial lowest label $(0,1)$ to $(-1,1)$ at the final coupling. This checks periodicity of the sector spectrum. It does not turn the two labels into the same wavefunction at fixed theta. Nor is the crossing a claim about a phase transition in interacting non-Abelian theory: it is an exact crossing in a specified free, fixed-magnetic-flux sector. The unconstrained Maxwell flux vacuum has all six integers zero.

2. Check a transformation with all four entries nonzero

Take $\tau_A=i$ and the matrix with rows $(2,1)$ and $(1,1)$. Transform the two labels $\gamma_1=(1,2)$ and $\gamma_2=(0,1)$. Verify both flux quadratic forms and their antisymmetric pairing. Does the transformation describe a beta function?

Hint 1

Use the charge map in (N17.11), not the fractional transformation on each charge.

Hint 2

The transformed coupling has imaginary part $1/2$. The pairing should remain $1$.

Solution

The determinant is $2-1=1$, and

$$\tau_A'=\frac{2i+1}{i+1}=\frac{3+i}{2}, \qquad \gamma_1'=(0,1),\quad \gamma_2'=(-1,1).$$

Initially the quadratic forms $|n+\tau_A m|^2/t$ are $5$ and $1$. Afterward they are

$$\frac{|(3+i)/2|^2}{1/2}=5,\qquad \frac{|(-1)+(3+i)/2|^2}{1/2}=1.$$

The initial pairing is $1\cdot1-2\cdot0=1$; the final one is $0\cdot1-1\cdot(-1)=1$. The box prefactors in (N17.6) consequently multiply equal energies before and after the map.

This is a discrete change of description and labels. No renormalization scale was varied, and no differential equation for a running coupling was introduced.

3. Try to change the electric unit

Rewrite precisely the same compact Maxwell theory using $\widehat a=2a$. Determine the new kinetic coupling $\widehat e$, theta coefficient $\widehat\theta$, holonomy period and permitted electric and magnetic labels. Show that the flux energy is unchanged. What goes wrong if one subsequently declares $\exp(i\oint\widehat a)$ to be the smallest allowed Wilson probe and allows every integer magnetic flux for $\widehat a$?

Hint 1

Substitute $f=\widehat f/2$ into the density. The old unit probe becomes $\exp(i\oint\widehat a/2)$.

Hint 2

The new holonomy has period $4\pi$. Its momentum eigenvalues are half-integers, while its fluxes in units of $2\pi$ are even integers.

Solution

Every quadratic term contains $f^2=\widehat f^2/4$, so

$$\widehat e^{\,2}=4e^2,\qquad \widehat\theta=\frac{\theta}{4},\qquad \widehat\tau_A=\frac{\tau_A}{4}.$$

But $\widehat q=2q$ has period $4\pi$, and $\exp(inq)=\exp(i(n/2)\widehat q)$. The unchanged theory therefore has

$$\widehat n=\frac n2,\qquad \widehat m=2m,\qquad (\widehat n,\widehat m)\in \tfrac12\mathbb Z\times2\mathbb Z.$$

In particular, this rewriting retains the lift to the original connection; it is not a sum over all ordinary unit-probe bundles for $\widehat a$. The energy test is

$$\frac{|\widehat n+\widehat\tau_A\widehat m|^2} {\operatorname{Im}\widehat\tau_A} =\frac{|\frac12(n+\tau_A m)|^2}{t/4} =\frac{|n+\tau_A m|^2}{t}.$$

The antisymmetric pairing is unchanged too. If one replaces the inherited labels by $\mathbb Z^2$ and discards the half-charge probe, the global data have changed: odd $\widehat m$ sectors have been added and half-integer $\widehat n$ sectors removed. Agreement of the local Maxwell density does not make that a mere rewriting. This is why a duality statement must specify charge units and allowed bundles together.